Variants of Turing Machines Question Computability and Complexity - - PDF document

Computability and Complexity

Lecture 2

Multitape Turing machines Nondeterministic Turing machines Enumerators Church-Turing Thesis

given by Jiri Srba

Lecture 2 Computability and Complexity 1/15

Variants of Turing Machines

Question What happens if we modify the definition of a Turing machine? Can we then possibly recognize more languages? Example: δ : Q × Γ → Q × Γ × {L, R, S} where S means “stay”. Answer No, the notion of a TM is robust. Hence no reasonable extension

• f a TM increases its power.

Example: “Stay” can be simulated in ordinary TM by two head movements (move right, move left).

Lecture 2 Computability and Complexity 2/15

Multitape Turing Machine

Definition (Multitape Turing Machine) A k-tape TM is a 7-tuple M = (Q, Σ, Γ, δ, q0, qaccept, qreject) where δ : Q × Γk → Q × Γk × {L, R}k and the rest is the same as before. Remark: 1-tape TM is exactly our original definition of TM. Configuration: a control state, plus the content of all k tapes together with the position of k heads. Initial configuration: input string written on the first tape, all

• ther tapes are empty (contain the blank symbols).

Computational step: all heads can move independently.

Lecture 2 Computability and Complexity 3/15

Equivalence of 1-tape and k-tape TM

Theorem Every k-tape TM is equivalent to 1-tape TM. Equivalent means that they recognize the same language. Corollary A language is recognizable iff it is recognized by some multitape Turing machine. Corollary A language is decidable iff it is recognized by some multitape Turing machine which is a decider.

Lecture 2 Computability and Complexity 4/15

Nondeterministic Turing Machine

Definition (Nondeterministic Turing Machine) A nondeterministic TM is a 7-tuple M = (Q, Σ, Γ, δ, q0, qaccept, qreject) where δ : Q × Γ → 2Q×Γ×{L,R} and the rest is the same as before. Remark: Every deterministic TM is also a nondeterministic TM. Configurations and Initial configuration as before. Computation tree: a tree of all configurations reachable from the initial one. The tree can have infinite branches! Acceptance Condition of a Nondeterm. TM A nondeterministic TM M accepts a string w if the computation tree for M and w contains at least one accepting configuration.

Lecture 2 Computability and Complexity 5/15

Equivalence of Deterministic and Nondeterministic TM

Theorem Every nondeterministic TM is equivalent to some deterministic TM. Equivalent means that they recognize the same language. Corollary A language is recognizable iff it is recognized by some nondeterministic Turing machine. Corollary A language is decidable iff it is recognized by some nondeterministic Turing machine which is a decider. A nondeterministic TM is a decider if for any given input every branch in the computation tree is finite (accepts or rejects).

Lecture 2 Computability and Complexity 6/15

Enumerators

Another terminology for recognizable languages is the term Recursively Enumerable languages. Why? Definition (Enumerator) An enumerator is a 2-tape Turing machine with a special control state called qprint. Definition (Language Generated by Enumerator) Let E be an enumerator. We run E on the empty string as input. The language of E, denoted by L(E), is the collection of all strings that are on the second tape whenever E is in the state qprint. Remark: If E does not terminate then L(E) may be infinite. Strings in L(E) may repeat and may be printed in arbitrary order.

Lecture 2 Computability and Complexity 7/15

Theorem about Enumerators

Theorem A language L is recognizable if and only if there exists an enumerator E that enumerates L, i.e. L(E) = L.

Lecture 2 Computability and Complexity 8/15

Proof (Enumerable = ⇒ Recognizable)

Every enumerable language L is recognizable: Let E be an enumerator for L. We construct a recognizer M for L. M = ” On input w:

• 1. Run E.
• 2. If w gets ever printed then M accepts,
• therwise continue running E in step 1.”

Lecture 2 Computability and Complexity 9/15

Proof (Recognizable = ⇒ Enumerable)

Every recognizable language L is enumerable: Let M be a recognizer for L. We construct an enumerator E for L. Let s1, s2, s3, . . . be all possible strings from Σ∗. E = ”

• 1. i:= 1;
• 2. Run M for i steps on each input s1, s2, . . . , si.
• 3. If M accepted any of the strings, print it
• 4. i:=i+1; goto step 2.”

This technique is called Dovetailing.

Lecture 2 Computability and Complexity 10/15

Hilbert’s Tenth Problem

In 1900 David Hilbert asked to find a mechanical way to check whether a polynomial (over several variables) has an integral root. Example: 6x3yz2 + 3xy2 − x3 − 10 has an integral root at x = 5, y = 3 and z = 0. Answer Nobody has found such an algorithm yet ... in fact, we know that this is impossible, and we can prove this!!! (Yuri Matijasevic’1970). We need a model of an algorithm to demonstrate such a proof.

Lecture 2 Computability and Complexity 11/15

Church-Turing Thesis

Models of an algorithm: 1936: Alan Turing came with Turing machine and Alonzo Church with λ-calculus. Turing machines and λ-calculus were shown equivalent. Many more models suggested: Kleene’s µ-recursive functions, Post-systems, Minsky machine ... all shown equivalent! Church-Turing Thesis Algorithms = Turing machines (informal notion) (formal, mathematical, concept) Facts: Church-Turing Thesis cannot be proved, BUT ... any C/Java/C++/C# program can be run on a TM, and nothing better than a Turing machine has been found so far.

Lecture 2 Computability and Complexity 12/15

Algorithmic Problems vs. Decidable Languages

How do algorithmic problems correspond to languages? Example: Hilbert’s Tenth Problem Can the problem whether a given polynomial has an integral root be algorithmically solved? Language Formulation of Hilbert’s Tenth Problem D def = {p | p is a polynomial with integral root } where p is the textual (string) encoding of the polynomial p. Is D a decidable language? Algorithmically solvable problems ≡ decidable languages.

Lecture 2 Computability and Complexity 13/15

Algorithmic Problems vs. Decidable Languages: Examples

Graph Connectivity “Is a given graph G connected?” corresponds to: Does G (encoding of G) belong to the language Lconnected

def

= {G ′ | G ′ is a graph and G ′ is connected } ? Acceptance Problem of a TM “Does a given TM M accept a string w?” corresponds to: Does M, w belong to the language ATM

def

= {M′, w′ | M′ is a TM and M′ accepts w′ } ? Facts: Lconnected is decidable, while ATM is undecidable!

Lecture 2 Computability and Complexity 14/15

Exam Questions

Equivalence of k-tape TM with 1-tape TM. Nondeterministic TM, definition, acceptance of a string, equivalence with ordinary TM. Enumerators, definition, equivalence with ordinary TM. Dovetailing technique. Church-Turing Thesis.

Lecture 2 Computability and Complexity 15/15